Problem 8.7. Modal analysis – Solution Manual

A one way slab, modelled as a beam with length L = 7.5 m, bending stiffness EI = 30×10⁶ Nm²/m and uniformly distributed mass m = 1200kg/m² is subjected to a uniformly distributed harmonic load: $\tilde{p} = p_{1} \sin (2 π f \times t), p_{1} = 293 N / m^{2}$ . The damping ratio is ξ = 2 %.

Determine the maximum acceleration and displacement from the steady-state solution using the modal analysis with the ABSSUM rule

a) The excited frequency is f = 2.5 Hz.

b) Assume that the excited frequency may vary between 2 and 3 Hz.

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Problem a)

Maximum displacement for f = 2.4 Hz, v_dyn[mm]=

Maximum acceleration for f = 2.4 Hz, a_dyn[m/s²]=

Problem b)

Maximum displacement for f = 3 Hz, v_dyn[mm]=

Maximum acceleration for f = 3 Hz, a_dyn[m/s²]=

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Steps

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Step 1. Give the first five eigen modes of a one way slab (beam).

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This image has an empty alt attribute; its file name is 8_7_1-1.jpg

Eq.(8-48)

$φ_{i} (x) = \sin (i \frac{π x}{L}), i = 1, 2, . . ., 5$

Step 2. Replace the load by the eigen mode series expansion.

Check eigen mode expanison of the loadHide eigen mode expanison of the load

The eigenmode series expansion of the load – since the eigenmodes are sinusoidal – is identical to its Fourier series expansion. The first five terms are:

See Figure 3.12 and Example 6.3.

$p_{1} = {\sum a_{i} \sin \frac{i π x}{L}, where}_{i = 1}^{5} a_{1} = \frac{4 p_{1}}{π} = 373.1, a_{2} = 0, a_{3} = \frac{4 p_{1}}{3 π} = 124.4, a_{4} = 0, a_{5} = \frac{4 p_{1}}{5 π} = 74.6$

Step 3. Calculate the static displacements due to the sinusoidal loads.

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See Example 3.6.

$v_{i} = \frac{1}{π} \frac{a_{i}}{E I} {(\frac{L}{i})}^{4} \sin \frac{πix}{L}, v_{1} = 403.9 \times 10^{- 6} m, v_{3} = 1.66 \times 10^{- 6} m, v_{5} = 0.129 \times 10^{- 6} m$

Step 4. Determine the eigenfrequencies.

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The response is determined for the three nonzero terms. The eigenfrequencies are:

Eqs.(8-47), (8-1)

$f_{i} = \frac{ω_{i}}{2 π} = \sqrt{\frac{π^{2} E I}{4 m L^{4}}} i^{2} : f_{1} = 4.415 \frac{1}{s}, f_{3} = 39.74 \frac{1}{s}, f_{5} = 110.4 \frac{1}{s}$

Problem a)

Step 5. Give the corresponding β-s and the dynamic amplification factors for f = 2.4 Hz.

Eqs.(8-22), (8-24)

$β_{i} = \frac{f}{f_{i}}, D_{δ i} = \frac{1}{\sqrt{{(1 - β_{i}^{2})}^{2} + {(2 ξ β_{i})}^{2}}}, D_{i} = β_{i}^{2} D_{δ i}$

i	1	3	5
β_i	0.5662	0.06291	0.02265
D_δi	1.471	1.004	1.001
D_i	0.4716	0.00397	0.000513

Step 6. Calculate the maximum dynamic displacement and the maximum acceleration.

The maximum dynamic displacement can be calculated as:

Eqs.(8-23), (8-73)

$v_{dyn} = |D_{δ 1} v_{1}| + |D_{δ 3} v_{3}| + |D_{δ 5} v_{5}| = (594.2 + 1.67 + 0.129) 10^{- 6} = 596.0 \times 10^{- 6} m = 0.596 m$

The maximum acceleration is

Eqs.(8-23), (8-73)

${\ddot{v}}_{dyn} = |D_{1} \frac{a_{1}}{m}| + |D_{3} \frac{a_{3}}{m}| + |D_{5} \frac{a_{5}}{m}| = 0.1466 + 0.0004118 + 0.000032 = 0.1471 \frac{m}{s^{2}}$

We may observe that the first term dominates and the contribution of the higher terms is less then 0.5%.

Problem b)

Step 5. Give the corresponding β-s and the dynamic amplification factors for f = 3 Hz.

When the excited frequency is closer to the eigenfrequency the responses are higher. For f = 3 Hz the β-s and the dynamic amplification factors are:

Eqs.(8-22), (8-24)

$β_{i} = \frac{f}{f_{i}}, D_{δ i} = \frac{1}{\sqrt{{(1 - β_{i}^{2})}^{2} + {(2 ξ β_{i})}^{2}}}, D_{i} = β_{i}^{2} D_{δ i}$

i	1	3	5
β_i	0.6794	0.07549	0.02718
D_δi	1.855	1.006	1.001
D_i	0.8564	0.00573	0.000739

Step 6. Calculate the maximum dynamic displacement and the maximum acceleration.

The maximum dynamic displacement can be calculated as:

Eqs.(8-23), (8-73)

$v_{dyn} = | D_{δ 1} v_{1} | + | D_{δ 3} v_{3} | + | D_{δ 5} v_{5} | = (749.3 + 1.67 + 0.129) 10^{- 6} = 751.1 \times 10^{- 6} m = 0.751 mm$

The maximum acceleration is ${\ddot{v}}_{d y n} = |D_{1} \frac{a_{1}}{m}| + |D_{3} \frac{a_{3}}{m}| + |D_{5} \frac{a_{5}}{m}| = 0.2662 + 0.000594 + 0.000046 = 0.2669 \frac{m}{s^{2}}$

Results

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The first five eigen modes of a one way slab (beam) are shown in the Figure.

This image has an empty alt attribute; its file name is 8_7_1-1.jpg

Eq.(8-48)

$φ_{i} (x) = \sin (i \frac{π x}{L}), i = 1, 2, . . ., 5$

The load is replaced by the eigenmode series expansion, which – since the eigenmodes are sinusoidal – is identical to its Fourier series expansion. The first five terms are:

See Figure 3.12 and Example 6.3

$p_{1} = \sum_{i = 1}^{5} a_{i} \sin \frac{i π x}{L}, where a_{1} = \frac{4 p_{1}}{π} = 373.1, a_{2} = 0, a_{3} = \frac{4 p_{1}}{3 π} = 124.4, a_{4} = 0; a_{5} = \frac{4 p_{1}}{5 π} = 74.6$

The static displacements due to the three sinusoidal loads are:

See Example 3.6.

$v_{i} = \frac{1}{π} \frac{a_{i}}{E I} {(\frac{L}{i})}^{4} \sin \frac{π i x}{L}, v_{1} = 403.9 \times 10^{- 6} m, v_{3} = 1.66 \times 10^{- 6} m, v_{5} = 0.129 \times 10^{- 6} m$

Now the response is determined for the three nonzero terms. The eigenfrequencies are:

Eqs.(8-47), (8-1)

$f_{i} = \frac{ω_{i}}{2 π} = \sqrt{\frac{π^{2} EI}{4 m L^{4}}} i^{2} : f_{1} = 4.415 \frac{1}{s}, f_{3} = 39.74 \frac{1}{s}, f_{5} = 110.4 \frac{1}{s}$

Problem a)

The corresponding β-s and the dynamic amplification factors for f = 2.4 Hz are:

Eqs.(8-22), (8-24)

$β_{i} = \frac{f}{f_{i}}, D_{δ i} = \frac{1}{\sqrt{{(1 - β_{i}^{2})}^{2} + {(2 ξ β_{i})}^{2}}}, D_{i} = β_{i}^{2} D_{δ i}$

i	1	3	5
β_i	0.5662	0.06291	0.02265
D_δi	1.471	1.004	1.001
D_i	0.4716	0.00397	0.000513

The maximum dynamic displacement can be calculated as:

Eqs.(8-23), (8-73)

$v_{dyn} = | D_{δ 1} v_{1} | + | D_{δ 3} v_{3} | + | D_{δ 5} v_{5} | = (594.2 + 1.67 + 0.129) 10^{- 6} = 596.0 \times 10^{- 6} m = 0.596 mm$

The maximum acceleration is

Eqs.(8-23), (8-73)

${\ddot{v}}_{d y n} = |D_{1} \frac{a_{1}}{m}| + |D_{3} \frac{a_{3}}{m}| + |D_{5} \frac{a_{5}}{m}| = 0.1466 + 0.0004118 + 0.000032 = 0.1471 \frac{m}{s^{2}}$

We may observe that the first term dominates and the contribution of the higher terms is less then 0.5%.

Problem b)

When the excited frequency is closer to the eigenfrequency the responses are higher. For f = 3 Hz the β-s and the dynamic amplification factors are:

Eqs.(8-22), (8-24)

$β_{i} = \frac{f}{f_{i}}, D_{δ i} = \frac{1}{\sqrt{{(1 - β_{i}^{2})}^{2} + {(2 ξ β_{i})}^{2}}}, D_{i} = β_{i}^{2} D_{δ i}$

i	1	3	5
β_i	0.6794	0.07549	0.02718
D_δi	1.855	1.006	1.001
D_i	0.8564	0.00573	0.000739

The maximum dynamic displacement can be calculated as:

Eqs.(8-23), (8-73)

$v_{dyn} = |D_{δ 1} v_{1}| + |D_{δ 3} v_{3}| + |D_{δ 5} v_{5}| = (749.3 + 1.67 + 0.129) 10^{- 6} = 751.1 \times 10^{- 6} m = 0.7511 mm$

The maximum acceleration is ${\ddot{v}}_{d y n} = |D_{1} \frac{a_{1}}{m}| + |D_{3} \frac{a_{3}}{m}| + |D_{5} \frac{a_{5}}{m}| = 0.2662 + 0.000594 + 0.000046 = 0.2669 \frac{m}{s^{2}}$